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nirvana33 [79]
3 years ago
5

Question 3 (5 points)

Chemistry
1 answer:
Daniel [21]3 years ago
8 0

Answer:

332.918g O2

Explanation:

I'm having some issues with the work however, your final answer should be 332.918g O2

Hope this helped!

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State the Constituents of the following alloys:magnalium and bronze​
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Magnalium: Magnesium and Aluminum

Bronze: Copper, Tin, Arsenic, Phosphorus, Aluminum, Manganese and Silicon (whichever you learned in class from those)
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Earth's Revolution question:
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Which sentence in a 5-paragraph essay states the essay's main idea?
dimaraw [331]
The main idea of an essay is usually portrayed in the thesis.
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What is the acceleration of the object’s motion? 0.5 m/s2 -0.5 m/s2 2 m/s2 -2 m/s2
djverab [1.8K]

Answer:  -2m/s2

Explanation:

Using the following equation ; acceleration = Change in velocity / time

i.e  a =  v - u / t

where  'a'  = acceleration

              v  = final velocity

              u  =  initial velocity

              t   =  time

Therefore;  from the graph we have acceleration to be, 0 - 6m/s / 3s  = -2m/s2

5 0
3 years ago
When 0.250 moles of KCl are added to 200.0 g of water in a constant pressure calorimeter a temperature change is observed. Given
777dan777 [17]

Explanation:

Upon dissolution of KCl heat is generated and temperature of the solution raises.

Therefore, heat generated by dissolving 0.25 moles of KCl will be as follows.

             17.24 kJ/mol \times 0.25 mol

                = 4.31 kJ

or,             = 4310 J      (as 1 kJ = 1000 J)

Mass of solution will be the sum of mass of water and mass of KCl.

       Mass of Solution = mass of water + (no. of moles of KCl × molar mass)

                                    = 200 g + (0.25 mol \times 54.5 g/mol)

                                    = 200 g + 13.625 g

                                    = 213.625 g

Relation between heat, mass and change in temperature is as follows.

                             Q = mC \Delta T

where,    C = specific heat of water = 4.184 J/g^{o}C

Therefore, putting the given values into the above formula as follows.

                     Q = mC \Delta T

            4310 J = 213.625 g \times 4.184 J/g^{o}C \times \Delta T      

              \Delta T = 4.82^{o}C

Thus, we can conclude that rise in temperature will be 4.82^{o}C.

6 0
3 years ago
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